IMO 2014/4

Points $$P$$ and $$Q$$ lie on side $$BC$$ of acute-angled triangle $$ABC$$ so that $$\angle PAB = \angle BCA$$ and $$\angle CAQ = \angle ABC$$. Points $$M$$ and $$N$$ lie on lines $$AP$$ and $$AQ$$ respectively, such that $$P$$ is the midpoint of $$AM$$, and $$Q$$ is the midpoint of $$AN$$. Prove that lines $$BM$$ and $$CN$$ intersect on the circumcircle of triangle $$ABC$$.

Note by Calvin Lin
3 years, 11 months ago

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I'll give a solution using complex numbers, hopefully no mistakes were made. The triangles QCA and ACB are similar, so (C-Q)/(A-Q)=[(C-A)/(B-A)]£, where £ refers to complex conjugation. Also, triangles ABC, PBA are similar, so (A-P)/(B-P)=[(C-A)/(B-A)]£ . But it is obvious that M-P=P-A and N-Q=Q-A, using these relations we get that (C-Q)/(N-Q)=(M-P)/(B-P), which means that the triangles CQN and MPB are similar. Now we observe that angle(ABM) +angle(ACN)=180 degrees, which means that ABCD is cyclic, so obviously D is on the circumcircle of the triangle ABC.

- 3 years, 11 months ago

Alternatively, you could note that $$\triangle BPM \sim \triangle CQN$$ and then perform a not too tedious trig bash.

- 3 years, 11 months ago

My friend said that he solved it with Barycentric coordinates in 5 minutes :O

- 3 years, 11 months ago